//
// Created by hitfy on 2018/5/1.
//

#include <bits/stdc++.h>

using namespace std;

int N;
int numbers[1000000];

int main(){
    scanf("%d", &N);
    int odd = 0, even = 0;
    for(int i = 0; i < N; i++){
        scanf("%d", &numbers[i]);
        if(numbers[i] % 2 == 1)
            odd++;
        else
            even++;
    }
    printf("%d", abs(odd - even));
    return 0;
}

/**
题目1 : Shortening Sequence
时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

There is an integer array A1, A2 ...AN. Each round you may choose two adjacent integers. If their sum is an odd number, the two adjacent integers can be deleted.

Can you work out the minimum length of the final array after elaborate deletions?
输入

The first line contains one integer N, indicating the length of the initial array.

The second line contains N integers, indicating A1, A2 ...AN.

For 30% of the data：1 ≤ N ≤ 10

For 60% of the data：1 ≤ N ≤ 1000

For 100% of the data：1 ≤ N ≤ 1000000, 0 ≤ Ai ≤ 1000000000
输出

One line with an integer indicating the minimum length of the final array.
样例提示

(1,2) (3,4) (4,5) are deleted.
样例输入

    7
    1 1 2 3 4 4 5

样例输出

    1
*/